# simple inequality problem with modulus

very sorry for the noobish post, I'm studying math on my own and lot of times im very confused about little things.

Here is my question, the original problem starts out like this: Solve the inequality $\vert x -2 \vert \geq \vert 2x -3\vert$

Here is the solved problem from my book

1. $\vert x -2 \vert \geq \vert 2x -3\vert$
2. $(x-2)^2 \geq (2x-3)^2$
3. $3x^2 - 8x + 5 \leq 0$
4. $(x-1)(3x-5) \leq 0$
5. $1 \leq x \leq 5/3$.

How does: $(x-1)(3x-5) \leq 0$. (Step 4)

Become solved as : $1 \leq x \leq 5/3$. (Step 5)

If it were separate wouldn't it be:

$$x \leq 1 \qquad \text{and} \qquad x \leq 5/3$$

So how does $x\leq 1$ become $1\leq x$??

Thanks so much!! =)

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oh didn't realize I had to use latex, sorry, there really should be a little latex ref for noobs in the faq!! –  gideon Oct 17 '10 at 10:53
Did you mean "inequality" in the title? =) –  Rasmus Oct 17 '10 at 10:55
sorry, corrected it, very frustrated with 10 days for my exam! =(, where oh where is the latex reference here! I see myself posting quite a few questions here! =P –  gideon Oct 17 '10 at 10:58
By the way, the easiest way to solve the original inequality is (in my opinion) simply to graph the two functions $f(x)=|x-2|$ and $g(x)=|2x-3|=2|x-3/2|$ and see for which $x$ the graph of $f$ lies above the graph of $g$. For your particular example you can easily see in the figure exactly where the graphs intersect; in general you might have to compute a little, but with the figure as a guide it's easier to avoid mistakes with signs and such. –  Hans Lundmark Oct 17 '10 at 13:33
Removed "modules" tag –  Byron Schmuland Oct 17 '10 at 15:26

So you want $(x-1) (3x-5) \leq 0$. So either one of the terms has to be positive and the other has to be negative.

• So if $x \geq 5/3$, then both the terms are positive and the product is positive.

• Similarly if $x \leq 1$, then both the terms are negative and their product is positive.

So observe that you don't want either of these things to happen, that is $x \notin (- \infty,1) \cup (5/3,\infty)$

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ok Im almost there, bear with me, if the first term is is negative so -x+1>0so 1> x and the other term positive so x < 3/5, but if (3x-5) is negative then the inequalities are 5/3 < x and x > -1, so the second set seem bogus, does that mean we accept the first? Thanks!=) –  gideon Oct 17 '10 at 10:52
I think you have edited the question. It didn't have mod when i saw it first –  anonymous Oct 17 '10 at 13:03
I didn't edit it someone else did, THIS -> |x−2|≥|2x−3| is the problem from my book, they square both sides and then factorize and come to this-->(x−1)(3x−5)≤0 then it turns into the answer, that part I do not get...this step ==>> (x−1)(3x−5)≤0. does not have the modulus in my book. Sorry I don't see the difference though! =S –  gideon Oct 17 '10 at 14:00
I edited the question again now.. –  gideon Oct 17 '10 at 14:08

Hint: In order for a product to be negative, one of the factors must be positive and the other one negative.

UPDATE: It's often helpful to draw a table such as this:

         x  |    1    5/3
------------+---------------
x-1  | -  0  +  +  +
3x-5  | -  -  -  0  +
------------+---------------
(x-1)(3x-5) | +  0  -  0  +


On each line you keep track of the signs of the individual factors in the intervals of interest. Then you construct the bottom line by using "minus times minus equals plus", "zero times anything equals zero", etc. In the bottom line you can then read off that $(x-1)(3x-5) \le 0$ is true exactly when $-1 \le x \le 5/3$.

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In the end here when you solve (x-1)(3x-5) <= 0 You are really finding the bounds of what x can be. In this case, the smallest number is 1, and the greatest number is 5/3. Therefore x has to be between 1 and 5/3. Which is where 1 <= x <= 5/3 comes from.

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thanks yea, I get it! =) thanks guys!! –  gideon Oct 17 '10 at 19:23